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QCD supplies the underlying strong interaction; Skyrmions provide a mesonic effective description of baryons, and quantized multi-Skyrmions can model nuclei. These are related descriptions at different scales, not three identical theories.

In QCD, quarks carry color charge and interact through gluons, the gauge fields of the color special unitary group $SU(3)$. Its Lagrangian density has the formAsymptotic freedom makes short-distance processes accessible through small-coupling expansions, but nuclear scales involve strongly coupled, confined dynamics. Observable hadrons are color singlets. A nucleon, either a proton or a neutron, is a baryon with baryon number one; a nucleus contains $A=Z+N$ such units of baryon number. Directly extracting all nuclear binding energies, spectra and interactions from QCD is difficult, motivating low-energy effective field theories that preserve its symmetries and relevant degrees of freedom.

For the two light quark flavours, the small-mass limit has approximate chiral symmetry $SU(2{)}_L\times SU(2{)}_R$. Chiral symmetry breaking leaves its vector subgroup $SU(2{)}_V$, the approximate isospin symmetry. The three pions are the associated Goldstone bosons in the massless limit and pseudo-Goldstone bosons when the light quark masses are retained. Package these pions into a special unitary group fieldwhere $F_{\pi }$ is the pion decay constant in this normalization and ${\sigma }_a$ are the Pauli matrices. The nonlinear sigma model is the leading two-derivative mesonic theory. The Skyrme model adds a specific four-derivative stabilizing interaction. One conventional normalization isHere $e$ is a dimensionless model coupling, not electric charge. The last term accounts for a common pion mass and preserves vector isospin. It vanishes in the chiral massless limit. This effective field theory uses color-singlet mesonic fields and does not resolve constituent quarks or gluons inside a baryon.

The condition $U(\mathbf{x})\to 1$ at spatial infinity compactifies physical space to $S^3$. Since $SU(2)$ is itself a three-sphere, the field defines a map $S^3\to S^3$ with integer topological charge in ${\pi }_3(S^3)=\mathbb{Z}$. This is identified with the topological baryon number in the Skyrme model:The associated topological current is identically conserved. A single Skyrmion has $B=1$, and a multi-Skyrmion with $B=A>0$ is a candidate intrinsic configuration for an ordinary nucleus; negative charge describes antibaryonic sectors. Integer topology prevents a smooth finite-energy unwinding into the classical vacuum, but it does not by itself guarantee a nonzero-size energy minimum.

The energetic reason for the Skyrme term is Derrick scaling. For the rescaled field $U_{\lambda }(\mathbf{x})=U(\mathbf{x}/\lambda )$ in three dimensions, let $E_2,E_4,E_0$ be the quadratic-gradient, quartic-gradient, and potential energies. Their scale dependence isThe two-derivative nonlinear sigma model alone can lower its static energy by shrinking. The positive quartic Skyrme term instead grows under shrinking, permitting a balance and a stable soliton size. Without the mass term, this balance gives $E_2=E_4$. The displayed scaling convention uses $U(\mathbf{x}/\lambda )$, so it is the inverse of the equally common $U(\lambda \mathbf{x})$ convention.

The connection with QCD is strengthened by large-Nc baryon scaling. Generalize the number of colors to $N_c$ while keeping $g_s^2{N}_c$ fixed. Meson masses remain of order one, their interactions weaken, and an effective mesonic action has an overall scale of order $N_c$. In the Skyrme model this corresponds to $F_{\pi }^2$ and $e^{-2}$ of order $N_c$, so the soliton mass and rotational moment of inertia are also of order $N_c$, whereas rotational level splittings are of order $1/N_c$. These are the expected baryon scaling properties of large-$N_c$ QCD. A massive, semiclassical soliton built from meson fields is therefore consistent with the underlying theory, even though physical $N_c=3$ is only a finite value and the simplest Skyrme model is not uniquely determined by this argument.

To represent a nucleon, a classical Skyrmion must be quantized. The unit Skyrmion hedgehog ansatz ties spatial rotations to isospin rotations. Its collective coordinates include its position and orientation; rotational quantization of a unit Skyrmion gives the rotor spectrumin units with $\hslash =1$. The Finkelstein-Rubinstein constraints impose the correct fermionic sign under a nontrivial configuration-space loop. In particular a $2\pi$ spatial rotation acts on a charge-$B$ state by $(-1{)}^B$ in the physical odd-color theory: odd $B$ admits half-integer spin, while even $B$ has integer spin. For $B=1$, the lowest allowed $J=I=1/2$ doublet represents the proton and neutron; the $J=I=3/2$ rotor state represents the Delta baryon resonance. A bosonic pion field can therefore describe fermionic baryons because the quantum wavefunction carries this nontrivial topological sign.

For nuclei, minimize the classical energy in a fixed baryon number sector, then quantize the permitted rotations, isospin rotations, and relevant vibrations or relative motions. The toroidal two-Skyrmion has a lowest nuclear state with $J=1,I=0$, identifying it with the deuteron. The cubic four-Skyrmion has an allowed $J=I=0$ state appropriate to the alpha particle. The rational map approximation for Skyrmions makes these intrinsic symmetries easier to construct, while collective-rotation constraints for a Skyrmion select allowed nuclear quantum numbers. A spin-zero state has rotationally invariant laboratory expectation values; a classical cubic intrinsic field should not be interpreted as a fixed cube visible in every orientation. Collective-coordinate quantization restores this distinction between intrinsic shape and a physical quantum state.

The nuclear force also has a mesonic interpretation. At large separation the tails of Skyrmions are weak pion fields; with nonzero mass their multipole falloff derives from derivatives of the Yukawa potential. Their interaction depends on relative orientation, and after quantization generates the familiar spin- and isospin-dependent pion-exchange structure of the nuclear force. Attractive channels allow several unit Skyrmions to form a lower-energy multi-Skyrmion. In nuclear language the positive nuclear binding energy is the difference between the separated nucleon masses and the mass of the quantized bound state, not just a count of topological units.

The limitations remain physical. The simplest Skyrme model retains only selected terms in a derivative expansion, and finite solitons probe gradients where omitted terms can matter. Its parameters require matching or calibration; predicted binding can be too strong, and masses, radii and spectra are not all fixed correctly by topology. Rotational quantization alone neglects quantum and vibrational corrections, especially when clustering or breakup channels are important. More general mesonic interactions, additional meson fields, and less restrictive classical ansätze can improve the description, but they introduce further low-energy information. The organizing relation is thereforeIt links underlying quark and gluon dynamics to a geometric, symmetry-based account of baryons and nuclei, while keeping the distinction between an effective approximation and a full derivation from QCD.

For a slowly evolving spherical star, use a Lagrangian enclosed mass coordinate $m$. Let $u$ be specific thermal internal energy and $D/Dt$ follow a mass element. The stellar energy balance equation isThe outward luminosity is energy transported through the mass shell, ${ϵ}_{\mathrm{nuc}}$ is nuclear rest-mass energy released per unit mass and time, and ${ϵ}_{\nu }$ is escaping neutrino power per unit mass. Define the gravothermal stellar energy generation by ${ϵ}_{\mathrm{grav}}=-Du/Dt-PD(1/\rho )/Dt$. At fixed composition it is $-TDS/Dt$ by the first law of thermodynamics. During composition changes the full internal-energy derivative must use the appropriate equation of state, with nuclear rest-mass changes counted only once. If a tabulated nuclear rate already subtracts reaction-neutrino energy, that loss must not be subtracted a second time in ${ϵ}_{\nu }$.

In a nearly steady main sequence star, stellar nuclear fusion supplies most of the luminosity. Before stable hydrogen ignition, Kelvin-Helmholtz contraction releases gravitational energy. In a gas-supported hydrostatic star the stellar virial theorem gives $U\simeq -\Omega /2$, so about half the change in gravitational binding energy heats the star and half is available for radiation. The associated timescale is $G{M}^2/(RL)$. After nuclear burning ends, a white dwarf can shine by losing stored thermal energy; crystallization releases latent heat, and composition separation can add gravitational energy. These are heat and gravitational reservoirs rather than sustained hydrogen fusion.

Stellar nuclear fusion is possible because lighter nuclei can combine into products with larger nuclear binding energy. The energy per reaction is $Q=(m_{\mathrm{initial}}-m_{\mathrm{final}})c^2$, including the relevant particles consistently. Although thermal energies are below the Coulomb barrier, quantum tunnelling permits fusion. The thermonuclear reaction rate averages a cross-section over the thermal relative-speed distribution:For nonresonant charged-particle reactions the energy weighting contains $\exp [-E/(kT)-\sqrt{E_G/E}]$; the compromise between the thermal tail and penetration produces the Gamow peak. A resonance can greatly enhance the rate. Density, abundances and temperature therefore all affect the stellar energy-generation rate, and power laws such as $T^4$ or $T^{16}$ are local approximations.

For the proton–proton chain, first produce deuterium and helium-3:The initial reaction is slow because it converts a proton to a neutron through the weak interaction. It is the bottleneck that allows long hydrogen-burning lifetimes. The alternative pep reaction $p+e^{-}+p\to {}^2\mathrm{H}+{\nu }_e$ also feeds the same chain. There are three principal proton-proton chain branches. In pp I, two helium-3 nuclei terminate the chain:In pp II, helium-3 first captures an existing helium-4 nucleus, then electron capture and proton capture finish the branch:In pp III, the beryllium-7 instead captures a proton:Relative branch weights change with temperature and composition. All branches convert four protons into a net helium-4 nucleus, with two weak conversions and two neutrinos, allowing for the electron consumed in pp II. After including positron annihilation, the atomic-mass energy budget is approximately $26.7\mathrm{MeV}$ per net helium-4 nucleus, but the deposited heat is smaller by the branch-dependent escaping-neutrino energy. Near ordinary low-mass main-sequence conditions, ${ϵ}_{pp}\propto \rho X_H^2{T}^4$ is a useful local approximation.

The CNO cycle provides an alternative catalyzed hydrogen burning route in hotter cores. The CNO-I cycle, also called the CN cycle, consists ofThe carbon seed is regenerated: the net reaction again consumes four protons and produces one helium-4 nucleus, two positrons and two neutrinos. The slow ${}^{14}\mathrm{N}(p,\gamma ){}^{15}\mathrm{O}$ reaction governs the ordinary cycle and causes nitrogen-14 to accumulate. The CNO-II cycle branches through ${}^{15}\mathrm{N}(p,\gamma ){}^{16}\mathrm{O}$, ${}^{16}\mathrm{O}(p,\gamma ){}^{17}\mathrm{F}$, ${}^{17}\mathrm{F}\to {}^{17}\mathrm{O}+e^{+}+{\nu }_e$, and ${}^{17}\mathrm{O}(p,\alpha ){}^{14}\mathrm{N}$, after which the main cycle continues. Thus CNO nuclei are catalytic in the closed cycles, although their relative abundances change. A local rate is ${ϵ}_{\mathrm{CNO}}\propto \rho X_H{X}_{\mathrm{CNO}}{T}^{\eta }$ with a much steeper exponent than the PP chain; the preceding homology problem specifies $\eta =16$. The concentration of this heating near the centre favors a convective core in hotter, more massive hydrogen-burning stars.

Once hydrogen is exhausted in a core, contraction can raise its temperature enough for core helium burning. The Triple-alpha process overcomes the absence of stable mass-5 and mass-8 nuclei by maintaining a tiny transient beryllium-8 population:The Hoyle state, an excited carbon-12 resonance near the three-alpha threshold, makes the second capture efficient enough despite beryllium-8's very short lifetime. The net conversion is $3{}^4\mathrm{He}\to {}^{12}\mathrm{C}$, releasing about $7.27\mathrm{MeV}$. Because three alpha particles are required, the specific rate scales as ${\rho }^2{Y}^3$; a schematic resonant rate isIts local temperature exponent is $-3+44/T_8$, of order forty near $T_8=1$. A degenerate core cannot expand promptly to regulate this increase, giving a helium flash; in a nondegenerate core the stellar thermostat permits stable helium burning. The competing ${}^{12}\mathrm{C}(\alpha ,\gamma ){}^{16}\mathrm{O}$ reaction determines much of the eventual carbon-oxygen mixture.

If the core becomes hot enough, carbon burning follows, for example through ${}^{12}\mathrm{C}+{}^{12}\mathrm{C}\to {}^{20}\mathrm{Ne}+\alpha$ and ${}^{12}\mathrm{C}+{}^{12}\mathrm{C}\to {}^{23}\mathrm{Na}+p$. Neon burning begins with photodisintegration, ${}^{20}\mathrm{Ne}+\gamma \to {}^{16}\mathrm{O}+\alpha$, followed by alpha capture such as ${}^{20}\mathrm{Ne}+\alpha \to {}^{24}\mathrm{Mg}+\gamma$. The first step consumes heat, but the combined rearrangement can release net energy. Oxygen burning includes ${}^{16}\mathrm{O}+{}^{16}\mathrm{O}\to {}^{28}\mathrm{Si}+\alpha$ and channels producing phosphorus and other nearby nuclei. At still higher temperatures, silicon burning proceeds through photodisintegration and particle captures in a reaction network approaching quasi-statistical equilibrium, producing iron-group nuclei. It is not simply direct silicon-plus-silicon fusion; the products depend on the electron fraction and weak-interaction timescale.

The increasing nuclear binding energy per nucleon supplies progressively less energy as products approach the iron group. Further fusion past that region cannot provide sustained net heat to support a core. Massive stars can therefore end with an unstable iron-group core; lower-mass stars do not attain all these burning stages and instead leave remnants such as carbon-oxygen white dwarfs. Heavy-element neutron captures can synthesize nuclei beyond the iron group without being the principal hydrostatic power source.

Finally, stellar neutrino energy loss competes with every heat source. Reaction neutrinos accompany the proton–proton and CNO chains. Hot or dense material also emits neutrino pairs through $e^{-}+e^{+}\to \nu +\bar{\nu }$, plasmon decay, $e^{-}+\gamma \to e^{-}+\nu +\bar{\nu }$, and electron-ion bremsstrahlung. These neutrinos usually escape, unlike the photons whose transport is diffusive; neutrino trapping requires much more extreme collapse conditions. Late burning has smaller fuel-energy reservoirs and strong neutrino cooling, so its duration is much shorter than core hydrogen burning. The surface luminosity is the escaping photon power, $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$, not the sum of photons and unobserved neutrino power. Radiative diffusion in a star, convection and electron conduction redistribute energy, while the energy-balance equation distinguishes true sources, losses and storage. Accretion or mass loss adds boundary energy and mechanical work when present, rather than changing the nuclear reaction budget.

The common energetic principle behind nuclear fission and nuclear fusion is the nuclear binding energy. If the products are more tightly bound in total than the reactants, their smaller rest mass releasesThe two binding-energy expressions assume that the same numbers of protons and neutrons are counted on both sides; emitted free nucleons have zero nuclear binding. The energy appears mainly as product kinetic energy, with radiation and subsequent decays contributing according to the reaction.

The binding energy per nucleon rises rapidly through light nuclei, has a broad maximum in the iron/nickel region, and decreases more slowly for heavier nuclei. Combining suitable light nuclei can therefore increase their total binding, whereas splitting a sufficiently heavy nucleus into moderately heavy fragments can do the same. This is an energy comparison, not a rule that every fusion or fission channel is exothermic: the actual nuclear masses fix the sign of $Q$.

In nuclear fission, the liquid-drop model illustrates the competition between attractive short-range nuclear cohesion and long-range proton repulsive Coulomb interaction. Deformation increases surface area and costs surface energy, but separating the charge can reduce nuclear Coulomb energy. Their competition creates a fission barrier. An excited nucleus may pass over the barrier; spontaneous fission can occur through quantum tunnelling. The resulting fragments are often excited and neutron-rich, so they may emit prompt neutrons and photons and undergo later beta decay. The typical total energy released by a heavy-nucleus fission is of order a few hundred MeV, rather than an electronic-scale energy. Its size follows from an approximately MeV-per-nucleon binding gain summed over a heavy nucleus. A neutron from one fission can induce another; a nuclear chain reaction is the repeated physical process, distinct from the energy-release condition for one event.

In nuclear fusion, the initial charged nuclei repel one another. Exothermicity does not remove this entrance Coulomb barrier. A reaction requires sufficient overlap of their wavefunctions, achieved by energetic encounters and quantum tunnelling. For nonresonant low-energy charged-particle fusion, a useful separation isThe astrophysical S factor isolates the nuclear part from much of the rapidly varying Coulomb penetration probability. In a thermal population, the competition between the decreasing high-energy population and increasing penetration produces the Gamow peak. This explains how fusion can proceed below the classical barrier without requiring the mean thermal energy to equal its height.

Both processes conserve total energy, charge, baryon number and angular momentum. Their very different entrance barriers and mass ranges explain why a favorable $Q$ value by itself does not specify a reaction rate. Short-range attraction gives approximately saturated nuclear binding, while surface, Coulomb, symmetry and pairing terms account for the broad trend and its isotope-dependent departures.

The nuclear binding energy is the separated-nucleon rest energy minus the bound-state rest energy. Consistent nuclear masses avoid atomic-electron bookkeeping:It is positive because the bound deuteron has lower total energy.

Smooth-model fission energy. Use natural units $c=1$, so the quoted nuclear masses are expressed as rest energies. Conservation of mass number and atomic number fixes the second fragment to $(A_2,Z_2)=((1-y)A,(1-y)Z)$. First omit nuclear pairing to compare a continuous range of $y$. Pairing can subsequently add ${\delta }_1+{\delta }_2-{\delta }_{\mathrm{parent}}$ for integer fragments; it has no smooth parity assignment for arbitrary real $y$.

In the parent-minus-daughters mass difference, the nucleon rest energy and volume terms cancel. The nuclear symmetry energy also cancels because both fragments have the same $Z/A$:The remaining liquid-drop fission energy isSplitting increases the surface energy cost, but reduces nuclear Coulomb energy. Both contributions are symmetric under $y\leftrightarrow 1-y$, so the symmetric split is stationary. Symmetry alone would not establish a maximum, since the two contributions compete.

For a full check, put $S=a_S{A}^{2/3}$, $x=a_C{Z}^2/(a_{S}A)$ and $t=y^{1/3}+(1-y{)}^{1/3}$. On $0\le y\le 1/2$, $t$ increases from $1$ to $2^{2/3}$. Expressing the two fractional-power sums through $t$ givesThe derivative can change sign only from negative to positive, so any interior stationary point in $t$ is a minimum. The largest value is therefore either the no-split endpoint $Q=0$ or the symmetric split. Thus symmetric fission maximizes the positive liquid-drop energy release: whenever this smooth model permits an exothermic split, its largest release occurs at $y=1/2$.

There is a qualification to the paper's unqualified wording. For arbitrary low $Z^2/A$, the symmetric split need not maximize $Q$; indeed it is a local minimum when $Z^2/A<(2/5)a_S/a_C=10$. The symmetric-maximum conclusion is valid in the heavy, fission-favorable regime relevant here, rather than a universal property of the formula.

At $y=1/2$,Requiring $Q_{\mathrm{sym}}>0$ gives

Barrier and fragment yields. Positive final-state Q value does not remove the fission barrier along a deformation path. Surface energy initially rises as the nucleus deforms; spontaneous fission then requires collective quantum tunnelling unless enough excitation is present. Its rate is exponentially sensitive to barrier height, width and collective inertia. Thus nuclei close to the exothermicity threshold can have immeasurably long spontaneous-fission mean lifetimes, while sufficiently heavier nuclei have smaller effective fission barriers or larger quantum tunnelling probabilities.

The fission-fragment mass distribution of low-excitation actinide fission is usually double peaked: complementary fragments have mass fractions roughly $0.4$ and $0.6$ rather than both exactly $1/2$. Nuclear shell model effects and the fission dynamics help favor asymmetric fragments. Maximizing the final smooth-model energy release is not a calculation of the fragment yield. The following yield curve is schematic, with no measured counts implied:

Capture excitation energies. Capturing a negligible-energy neutron forms a compound nucleus with the rest energy of the target plus the free neutron. Relative to the compound ground state, the neutron-capture excitation energy is thereforewhere $B$ is positive nuclear binding energy and $S_n$ the compound nucleus's neutron separation energy. The free neutron rest energy cancels in the binding-energy difference. With the supplied semi-empirical mass formula,At $Z=92$, uranium-235 has odd $A$, while uranium-236 is even-even. The pairing contribution is $+33.5(236{)}^{-3/4}=+0.55637\mathrm{MeV}$. By contrast uranium-238 is even-even and uranium-239 has odd $A$, giving $-33.5(238{)}^{-3/4}=-0.55286\mathrm{MeV}$. The individual contributions areThus the model predictsOf the $1.48305\mathrm{MeV}$ predicted difference, $1.10922\mathrm{MeV}$ comes from the change in nuclear pairing. It is the primary reason the even-even capture product is more strongly excited; the smooth terms account for the remaining $0.37383\mathrm{MeV}$.

Thermal versus fast-neutron fission. The last comparison uses the supplied observed capture excitations, $6.5$ and $4.8\mathrm{MeV}$, rather than replacing them by the preceding smooth-model estimates. Interpreting the quoted $6.2$ and $6.6\mathrm{MeV}$ activation energies as the barriers in the respective capture channels givesA simple neutron-induced fission threshold estimate for the second channel requires about $6.6-4.8=1.8\mathrm{MeV}$ additional centre-of-mass kinetic energy. Sub-barrier tunnelling is possible, but it does not give the large thermal-neutron fission probability of the first channel.

The barrier labels in the original PDF need care: it assigns them to ${}^{235}\mathrm{U}$ and ${}^{238}\mathrm{U}$, whereas the states formed after capture are ${}^{236}{\mathrm{U}}^{\ast }$ and ${}^{239}{\mathrm{U}}^{\ast }$. The comparison above uses the intended capture-channel interpretation. If the quoted numbers are literally barriers of the uncaptured target nuclei, the compound-nucleus barriers are missing and this numerical comparison does not follow from those data alone.

In a thermal reactor, neutron moderation slows fission neutrons so the large low-energy fission cross section of uranium-235 can support a nuclear chain reaction. Fuel composition, moderator choice and geometry must balance production against capture and leakage; capture resonances of uranium-238 during slowing are important losses from that neutron population. The reaction rate per target nucleus depends on the spectrum through $\int \varphi (E){\sigma }_f(E)dE$, where $\varphi (E)$ is differential neutron flux per unit energy. A fast spectrum can access the uranium-238 fission channel and changes this balance. Also, uranium-238 is a fertile material: capture followed by two beta-minus decays can produce plutonium-239, a fissile material, so capture can contribute to breeding even when it causes no immediate fission.

Dating principle. Atmospheric cosmic rays produce secondary neutrons, which can generate carbon-14 through ${}^{14}\mathrm{N}(n,p){}^{14}\mathrm{C}$. Carbon then enters atmospheric carbon dioxide and living organisms. While an organism exchanges carbon with its environment, its carbon-14 to carbon-12 ratio approximately tracks that reservoir. After exchange stops, surviving carbon-14 undergoes beta-minus decay, whereas the stable carbon remains. On the closed-sample, constant-initial-ratio model, radioactive decay givesThus radiocarbon dating infers elapsed time from the loss of an isotope ratio or of radioactive activity per fixed carbon mass. It dates cessation of carbon exchange, which need not equal the date an object was manufactured or buried. Actual calendar dating also requires radiocarbon calibration and control of contamination and reservoir differences.

Mean lifetime. For exponential decay, the half-life obeys $e^{-\lambda T_{1/2}}=1/2$, while the mean lifetime is $\tau =1/\lambda$. HenceUsing $1\mathrm{yr}=365.25\times 86400\mathrm{s}$ gives $\lambda =3.83325\times 10^{-12}{\mathrm{s}}^{-1}$ and $\tau =2.60875\times 10^{11}\mathrm{s}$.

The mass minimum and beta stability. Treat the supplied masses as nuclear rest energies in $\mathrm{MeV}$, and omit nuclear pairing as instructed. At fixed mass number $A$, the volume and surface terms of the semi-empirical mass formula are constant. Expanding the remaining terms gives a convex nuclear mass parabola,The continuous minimum in atomic number is thereforeAt $A=14$, the quadratic coefficient is $7.01302\mathrm{MeV}$ and $Z_{\ast }=6.79450$. The nearest integer is $Z=7$; the nuclear isobar of minimum mass is nitrogen-14. Maximizing nuclear binding energy alone would omit the neutron-proton rest-mass difference; its continuous optimum is slightly different, but also selects integer $Z=7$ here.

The values most useful for checking neighboring decay thresholds areConsequently carbon-14 can lower its nuclear mass by changing a neutron into a proton:For bare nuclear masses, the Q value must include the emitted electron mass. Taking $m_e{c}^2=0.511\mathrm{MeV}$,The opposite proton-to-neutron conversion would lead to $Z=5$, whose nuclear mass is $18.1567\mathrm{MeV}$ higher than that of carbon-14. Neither beta-plus decay nor electron capture can bridge that gap by supplying an electron-mass correction. For the nitrogen-14 daughter, both neighboring nuclear masses are higher: $Z=6$ by $4.13070\mathrm{MeV}$ and $Z=8$ by $9.89535\mathrm{MeV}$. Even electron capture to $Z=6$ has $Q=0.511-4.13070<0$. Thus the daughter has no further energetically allowed single beta decay in this model; physical nitrogen-14 is stable.

These threshold numbers belong to the supplied smooth semi-empirical mass formula, not to an accurate prediction of the carbon-14 electron spectrum or its long half-life. Semi-empirical mass errors can change beta-decay thresholds. Also, the instruction to omit nuclear pairing is a model simplification: even $A=14$ does not make physical pairing vanish, and even-even carbon and odd-odd nitrogen need not have the same pairing correction. No omitted pairing coefficient is guessed here.

Activity of modern carbon. Neglect the small contribution of other stable carbon isotopes when counting atoms in a gram. The Avogadro constant givesThe expected radioactive activity of that one gram isor approximately $15.0$ decays per minute per gram of carbon. A becquerel is one decay per second; the normalization is per gram of carbon, rather than per gram of the entire organic sample.

Artefact age. Its remaining activity fraction is $0.13/0.25008=0.51983$. Inverting the exponential decay law,This is the radiocarbon age on the supplied constant-initial-activity, $5730$-year half-life model. It is not a calibrated calendar date; the sample's historical initial ratio and any later carbon exchange are not supplied.

There are three free-particle rest-energy terms and five nuclear corrections. In the paper's natural units, their meanings are:

The sign convention is significant: the quantity labeled $E_{\mathrm{bind}}$ in the supplied mass formula is added to the free-particle masses and is normally negative for a bound nucleus. It is the negative of the conventional positive nuclear binding energy $B$:The coefficients must therefore retain their printed signs; using positive $B$ while still adding it to the masses would be inconsistent.

The right panel of the figure in solution has luminosity increasing upwards and effective temperature decreasing to the right. An old globular cluster contains a low-mass main sequence, a main-sequence turnoff, subgiants, a red giant branch, a horizontal branch, an Asymptotic giant branch and a faint hot white dwarf sequence. Its high-mass main sequence is absent because those stars have evolved.

If a fraction $f_c=0.1$ of the star participates in hydrogen burning, with initial hydrogen mass fraction $X=0.7$, the available nuclear binding energy is $0.007f_{c}XM{c}^2$. Assuming a roughly constant luminosity and coeval stars,Taking the quoted core fraction as hydrogen fuel itself, $f_{c}X=0.1$, gives $1.98\times 10^{10}\mathrm{yr}$. Stating the fuel convention matters: burning the whole stellar mass overestimates the age by about an order of magnitude. Evolving luminosity and detailed stellar structure are needed for an accurate age.

The right panel of the figure in solution has luminosity increasing upwards and effective temperature decreasing to the right. An old globular cluster contains a low-mass main sequence, a main-sequence turnoff, subgiants, a red giant branch, a horizontal branch, an Asymptotic giant branch and a faint hot white dwarf sequence. Its high-mass main sequence is absent because those stars have evolved.

If a fraction $f_c=0.1$ of the star participates in hydrogen burning, with initial hydrogen mass fraction $X=0.7$, the available nuclear binding energy is $0.007f_{c}XM{c}^2$. Assuming a roughly constant luminosity and coeval stars,Taking the quoted core fraction as hydrogen fuel itself, $f_{c}X=0.1$, gives $1.98\times 10^{10}\mathrm{yr}$. Stating the fuel convention matters: burning the whole stellar mass overestimates the age by about an order of magnitude. Evolving luminosity and detailed stellar structure are needed for an accurate age.

The first step of the proton–proton chain isHere $d$ is the deuteron. A weak charged current converts one proton into a neutron, while the resulting proton–neutron pair becomes bound. The Q value follows from the difference of initial and final rest energies:This includes the emitted Positron's rest energy in the final-state mass and excludes its subsequent electron-positron annihilation. The nuclear binding energy of the deuteron makes this reaction possible although the isolated conversion $p\to n+e^{+}+{\nu }_e$ is energetically forbidden. The released energy is shared by final-state kinetic energies, including the escaping neutrino.

Interpret the displayed semi-empirical mass formula as an equation for mass energy, using $c=1$ for the mass terms. Its seven contributions are as follows.

The term $Z(m_p+m_e)$ is the rest energy of the $Z$ protons and $Z$ atomic electrons; $(A-Z)m_n$ is the rest energy of the neutrons. Including $Z{m}_e$ makes this an atomic-mass approximation. Small atomic electronic binding energies are neglected.

The volume contribution $-a_{v}A$ represents the short-range attractive strong interaction. Saturation gives each nucleon roughly a fixed number of neighbors, so the nuclear binding energy is approximately proportional to $A$; binding lowers the mass. The surface contribution $+a_s{A}^{2/3}$ restores the binding missing from nucleons near the surface. Approximately constant nuclear density gives radius proportional to $A^{1/3}$ and area proportional to $A^{2/3}$, explaining that power and the positive mass correction.

The Coulomb potential energy contribution $+a_c{Z}^2/A^{1/3}$ is positive because proton charges repel. A sphere's electrostatic self-energy scales as $Z^2/R$, and $R\propto A^{1/3}$. The more refined proton-pair count $Z(Z-1)$ differs from the $Z^2$ approximation explicitly used here.

The nuclear symmetry energy contribution $+a_a(Z-A/2{)}^2/A$ penalizes unequal neutron and proton populations. By the Pauli exclusion principle, filling one species' higher-momentum states is energetically more expensive than balancing the two Fermi seas; nuclear interactions also contribute to the empirical coefficient. The stated convention is equivalent to $23.285(N-Z{)}^2/A$ MeV, since $Z-A/2=(Z-N)/2$. It must not be combined with the usual coefficient near $23$ MeV without this factor of four.

Finally, nuclear pairing favors paired protons and paired neutrons. In the mass formula,The signs reverse when this correction is written as a contribution to positive nuclear binding energy. This smooth liquid-drop model omits detailed nuclear shell corrections.

The bookkeeping reaction isBoth mass number and electric charge are conserved: $236=92+140+4$ and $92=37+55$. Express each atomic mass as the sum of its constituent rest masses minus its nuclear binding energy. The total constituent masses, including the atomic-electron masses, cancel between the two sides. Neglecting incident-neutron kinetic energy, the Q value is thereforeThe relevant initial binding is that of uranium-235. Introducing uranium-236 as an intermediate state is possible only if the neutron-capture energy is also accounted for.

In the stated coefficient convention,Uranium-235 has odd $A$, hence ${\delta }_B=0$. Both products are odd–odd: rubidium has $(Z,N)=(37,55)$ and caesium has $(55,85)$. Each therefore has ${\delta }_B=-12/\sqrt{A}$ MeV. Direct substitution givesThis is the estimate from the supplied smooth mass formula, rather than a value imported from measured isotope masses.

For $100$ MW recovered as the stated fission energy, the annual energy output is $10^8(365.25)(86400)$ J. ThusEach event consumes one uranium-235 atom. With Avogadro constant $N_A=6.02214076\times 10^{23}$ mol${}^{-1}$ and molar mass approximated as $235$ g mol${}^{-1}$,This uses the stated power directly as recovered fission power; an unspecified conversion efficiency or additional heat-loss correction cannot be inferred from it.

The semi-empirical mass formula approximates nuclear binding energy using volume, surface, electrostatic, neutron-proton asymmetry, and nuclear pairing contributions. At fixed mass number, its quadratic dependence on atomic number gives a nuclear mass parabola. It describes broad trends and does not resolve individual nuclear shell effects.

At very high temperatures, photodisintegration and proton, neutron and alpha captures rearrange silicon-group material through networks approaching quasi-statistical equilibrium. The result is an iron-group mixture determined by electron fraction and weak-reaction timescales. Direct silicon-plus-silicon fusion is not the usual mechanism. The approach to maximal nuclear binding energy per nucleon ends sustained hydrostatic fusion power.